[H simulation] lc68. Align text left and right (analog + classic)

1. Source of the subject

Link: 68. Align text left and right

2. Topic analysis

This simulation problem of string processing is quite common. There are three situations in which the topic is abstracted, and the first line of the note is the problem analysis. This is much more valuable than 65 questions...

This kind of simulation writing can also exercise one's engineering ability and design ability to a large extent.

• Time complexity : $O\left(n\right)$。
• Space complexity : $O\left(n\right)$

Code:

class Solution {
    
    
public:
    vector<string> fullJustify(vector<string>& words, int maxWidth) {
    
    
        vector<string> res;
        for (int i = 0; i < words.size(); i ++ ) {
    
    
            int j = i + 1;   // 顺序枚举所有单词。i 是当前行起始单词，j 是当前行末尾单词。构成单词区间
            int len = words[i].size();      // 存储当前行的单词长度。包含了相邻单词必带的一个空格

            // 当前行能放多少单词，当前枚举单词的长度+空格+下一个单词长度在限值内，则考虑下一个单词
            while (j < words.size() && len + 1 + words[j].size() <= maxWidth) 
                len += 1 + words[j ++ ].size();

            string line;            // 组织处理这一行的单词
            if (j == words.size() || j == i + 1) {
    
      // 如果 j 是最后一行或者这一行只放一个单词，则左对齐
                line += words[i];                  // 第 i 个单词的长度
                for (int k = i + 1; k < j; k ++ ) line += ' ' + words[k];  // 先组织成一个空格的左对齐形式
                while (line.size() < maxWidth) line += ' ';     // 补充末尾的连续空格
            } else {
    
        // 左右对齐
                // cnt 是单词间的待分配空格的次数，为两单词之间的间隔数
                // r   是剩余未填充进去的空格数
                // len 是包含了本行相邻单词的 1 个空格的单词长度。故剩余的未填充的空格数量得再补充上 cnt 个
                int cnt = j - i - 1, r = maxWidth - len + cnt;  
                line += words[i];  
                int k = 0;
                // 先填充空格，再填充单词，左边比右边多 1 的情况总共需要分配 r % cnt 个
                while (k < r % cnt) line += string(r / cnt + 1, ' ') + words[i + k + 1], k ++ ;
                // k 最多这样分配 cnt 次这样的空格。左边的多 1 的情况分配完毕，处理右边少的情况即可
                while (k < cnt) line += string(r / cnt, ' ') + words[i + k + 1], k ++ ;
            }
            res.push_back(line);
            i = j - 1;
        }
        return res;
    }
};